Minimizing a functional subject to boundary conditions

boundary value problemcalculus-of-variationsconstraintseuler-lagrange-equationvariational-analysis

I want to find a smooth function $y : [0,1] \to \mathbb{R}$ that minimizes

$$
S = \int_0^1 (y(x)y'''(x) + 3 y'(x) y''(x))^2 dx
$$
subject to the constraints / boundary conditions: (i) $y(0)=1$, (ii) $y(1)=0$, (iii) $y'(0)=0$, (iv) $y'(1)=0$.

What is the right approach for this? If I solve the unconstrained Euler-Poisson equation I get the square root of a fifth order polynomial, but this solution cannot satisfy all four constraints.

I understand how to implement constraints via Lagrange multipliers if the constraints are themselves differential equations, but is there a similar way to implement these boundary conditions?

(This is probably a duplicate of some sort, but I couldn't find the answer, mainly, I assume, because I don't know the right keyword to search for.)

Edit: As mentioned in the comments, I might be happy knowing just the minimal value for $S$, even if I don't find the explicit form of the minimizing $y(x)$.

With a mix of partial integration, Cauchy-Schwarz, and Taylor's theorem, I managed to show that $S \geq 1$ must hold, which is something already.

Best Answer

TL;DR: It is potentially problematic to assume a 6th-order Euler-Poisson (EP) equation in a higher-order variational problem without adequate boundary conditions (BCs).

  1. OP's functional can be rewritten as $$S[z]~:=~\frac{1}{4}\int_0^1 \!dx~ (z^{\prime\prime\prime}(x))^2, \tag{1}$$ where we have introduced a new variable $$z~=~y^2\tag{2}$$ which must satisfy an inequality $$ z(x)~\geq~0,\tag{3}$$ which one should check in the end.

  2. OP's BCs for $y$ $$ y(0)~=~1, \quad y^{\prime}(0)~=~0, \quad y(1)~=~0, \quad y^{\prime}(1)~=~0, \tag{4}$$ clearly imply the corresponding BCs for $z$ $$ z(0)~=~1, \quad z^{\prime}(0)~=~0, \quad z(1)~=~0, \quad z^{\prime}(1)~=~0. \tag{5}$$ Conversely, to ensure the last BC in eq. (4) [for a sufficiently nice function $y$] we need to impose a 5th BC $$ z^{\prime\prime}(1)~=~0, \tag{6}$$ as explained in Gonçalo's answer.

  3. If we vary infinitesimally OP's functional, we calculate $$ \begin{align}2\delta S ~\stackrel{(1)}{=}~~&\int_0^1 \!dx~ z^{\prime\prime\prime}(x)~\delta z^{\prime\prime\prime}(x)\cr ~=~~&\int_0^1 \!dx\frac{d}{dx}\left(z^{\prime\prime\prime}(x)~\delta z^{\prime\prime}(x)\right)-\int_0^1 \!dx~ z^{(4)}(x)~\delta z^{\prime\prime} \cr ~\stackrel{(5)+(6)}{=}&-z^{\prime\prime\prime}(0)~\delta z^{\prime\prime}(0)-\int_0^1 \!dx~ z^{(6)}(x)~\delta z. \end{align}\tag{7} $$

  4. Now for the 6th-order EP equation $$ z^{(6)}(x)~=~0. \tag{8}$$ to be a stationary solution, we need a 6th BC $$z^{\prime\prime\prime}(0)~=~0.\tag{9} $$

  5. One may check that the minimizing 5th-order polynomial found in Gonçalo's answer happens to satisfy the 6th BC (9) as well, so it is indeed a stationary solution!

    This is not a coincidence: When we minimized the functional (1) for 5th-order polynomials satisfying EP eq. (8) and BCs (5) and (6), then eq. (7) conversely implies the 6th BC (9).

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